Throwing sandbags out of air balloon

[mathmaniac1]

An air balloon of mass M descending down with acceleration 'a'.A mass m is dropped out of it and the acceleration becomes a upwards.Find m in terms of M,a...

The right answer is 2Ma/(g+a)

The explanation I was given:
The mass out of the balloon now has a+g acceleration.
(How could it be when g was already a part of a?)
The 'action'=m(a+g)
Reaction=Change in force=2Ma-(-2Ma)=2Ma
2Ma=m(a+g)---->Newton's third law.
(Wait,final force=(M-m)a,why not?could it be that it was just ignored?)
(I don't get how they can be identified as action and reaction forces)
m=2Ma/(g+a)

Please help.

 

[Nono713]

Here's a derivation that you might find more intuitive (to be honest I also find the explanation given very poor and handwavy, even though it may be longer the proof below might help you see what's going on and why $g$ is taken into account).

First, assume that the engine which is imparting an acceleration to the balloon (i.e. helium lift or whatever) produces constant thrust upwards, denoted $F_\mathrm{thrust}$, and that the gravitational acceleration is downwards and is denoted $g$ (and is positive just for consistency with the question's sign conventions). The only assumption is that throwing the sandbag out does not change this thrust, which is reasonable and apparently correct (since sandbags generally do not interact with the engine, just dead weight).

Before the sandbag is released, the balloon has two forces acting on it, with a net force of:

$$F_i = F_\mathrm{thrust} - Mg$$

After the sandbag is released, the balloon still has the same thrust, but the gravitational force is less:

$$F_f = F_\mathrm{thrust} - (M - m)g$$

And so we know that the following holds:

$$F_\mathrm{thrust} = F_i + Mg = F_f + (M - m)g$$

But we also know from the question that $F_i = -Ma$, and $F_f = (M - m)a$. Substituting:

$$-Ma + Mg = (M - m)a + (M - m)g$$

$$M(g - a) = (M - m)(g + a)$$

And solving for $m$, the unknown mass:

$$M \frac{g - a}{g + a} = M - m$$

$$m = M - M \frac{g - a}{g + a} = M\left (1 - \frac{g - a}{g + a} \right ) = M \left ( \frac{g + a}{g + a} - \frac{g - a}{g + a} \right ) = M \left ( \frac{2a}{g + a} \right ) = \frac{2Ma}{g + a}$$

$$\blacksquare$$

(I'm actually shocked I got the derivation right so quickly, I'm usually crap at Newtonian physics. I guess playing KSP every now and then really does help :p).

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